EXERCISE 2
Q1 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 1
Evaluate:
(i) 2-3
Sol :
Using property a−n=1an
⇒123=12×12×12
⇒12×2×2=18
(ii) 60
Sol :
Using property: a0=1
⇒60=1
(iii) (-3)-2
Sol :
Using property a−n=1an
⇒1(−3)2=1−3×1−3
⇒1−3×−3=19
(iv) 85 . 8-5
Sol :
Using property am×an=a(m+n)
⇒85 . 8-5 or
⇒85 × 8-5
⇒8(5-5)
⇒80
⇒1
(v) 12−2
Sol :
Using property 1a=a−1
⇒2(-2)×-1
⇒22
⇒4
(vi) 3(8-1)0
Sol :
Using property: a0=1
⇒3×1
⇒3
(vii) 12−1+13−1
Sol :
Using property : 1a−1=a
⇒2+3
⇒5
(viii) 7x0
Sol :
Using property: a0=1
⇒7×1
⇒7
(ix) a2 ÷ a-2
Sol :
Using property : a−n=1an
⇒a2÷1a2
⇒a2×a21
⇒a2×a2
Using property : an×am=a(n+m)
⇒a4
(x) y−6y4
Sol :
Using property : a−n=1a
⇒1(y4)×(y6)
Using property: am×an=a(m+n)
⇒1y(6+4)=1y(10) or
⇒y(-10)
(xi) (x -3)0
Sol :
Using property: (am)n=a(m×n)
⇒x0
⇒1
(xii) 60 + 6 -1 + 3 -1
Sol :
Using property: a0=1
⇒1+6 -1 + 3 -1
Using property: a−1=1a
⇒1+16+13
L.C.M of 6 and 3 is =2×3=6
26,333,31,1
⇒1×6+1×1+1×26=6+1+26
or 96=32
(xiii) 1a0+b0
Sol :
Using property: a0=1
⇒11+1=12
(xiv) (3x2)3
Sol :
Using property: (an)m=a(m×n)
⇒ 3(3)× x(2×3)
⇒ 27× x6
⇒ 27x6
(xv) [(57)−6]−0
Sol :
Using property: (am)n=a(m×n)
⇒(57)−6×−0
⇒(57)0
Using property : a0=1
⇒1
Q2 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 2
Simplify:
(i) x5 . x3
Sol :
Using property : am×an=a(m+n)
⇒x(5+3)
⇒x(8)
(ii) x8x3
Sol :
Using property : 1an=a−n
⇒x8 × x(-3)
Using property : am×an=a(m+n)
⇒x(8-3)
⇒x5
(iii) (x3)2
Sol :
Using property : am×an=a(m+n)
⇒x(3×2)
⇒x6
(iv) (3x2)3
Sol :
Using property : am×an=a(m+n)
⇒33×(x2)3
⇒27×x6
⇒27x6
(v) (5x-3yz2)-2
Sol :
Using property : a−n=1an
⇒1(5x−3yz2)2
⇒152×x−3×2×y2z2×2
⇒125×x−6×y2×z4
⇒x625y2z4
Q3 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 3
Write the option that indicates the correct meaning of the given expression
(i) 4n3 means
(a) 4 times n + n + n
(b) 4n . 4n . 4n
(c) 4 . n . n . n
Sol :
⇒4n3
⇒4.n.n.n or (4×n×n×n)
Option (c) is correct
(ii) 6x3y2 means
(a) 6 . xy . xy . xy . xy . xy
(b) (6 . x. x. x . y)(7. x. x. x . y)
(c) 6 . x . x . x . y .y
Sol :
Option (c) is correct
Q4 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 4
Simplify:
(i) (a0b−2)52a−1
Sol :
⇒(a0×5b−2×5)2a−1
⇒(a0b−10)2a−1
Using property: 1a−1=a
⇒a0×a12b10
Using property: a0=1
⇒a12b10
(ii) 4a−3b02a2b−1
Sol :
⇒⧸4a−3b0⧸2a2b−1
Using property: 1a=a−1
⇒2a-3×b0× a-2 × b+1
Using property: an×am=a(m+n)
⇒2a-3-2×b0+1
⇒2a-5×b1
⇒2ba5
(iii) 2x2y−22−1x2y2
Sol :
Using property : 1a−1=a1
⇒(2x2y-2)(21x-2y-2)
⇒(2×2)x2-2y-2-2
⇒4x0y-4
Using property : a0=1
⇒4y-4
⇒4y4
(iv) (t−4)3(t3)−4
Sol :
Using property: (am)n=a(m×n)
⇒t−12t−12
⇒1
(v) 4k2(4-1k+4k-2)
Sol :
⇒4k2(4-1k)+4k2(4k-2)
using property: 1a=a−1
⇒4k2×k4+4k2×4k2
⇒k2+11+16k2k2
⇒k3+16
(vi) (3x−22y−1)−2
Sol :
⇒(3x−2)−2(2y−1)−2
Using property : (am)n=a(m×n)
⇒3−2x4)2−2y2
⇒9−1x4)4−1y2
⇒4x4)9y2
(vii) [3−1(−2)−2]−2
Sol :
⇒(3−1)−2{(−2)−2}−2
Using property : (am)n=a(m×n)
⇒32(−2)4=916
Q5 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 5
Express each of the following as a power of 2 : 8 , 8x , 16x+3
Sol :
⇒8=2×2×2=23
⇒8x=(2×2×2)x=23x
⇒16x+3=(2×2×2×2)x+3
Q6 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 6
If a = 2m and b = 2{m+1} , show that 8a3b2=2m+1
Sol :
⇒8a3b2
⇒8(2m)3(2m+1)2
Using property : (am)n=a(m×n)
⇒23×23m22m+2
Using property : a(m×n)=(am)n
⇒23×22m+m22m+22
⇒23×22m×2m22m×22
⇒23×⧸(22m)×2m⧸(22m)×22
⇒2×2m
Using property : am×an=a(m+n)
⇒2m+1
Hence proved
Q7 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 7
If x = 2k and y = 2k+3 , what is the value of xy ?
Sol :
⇒xy
⇒2k2k+3
Using property: a(m+n)=am×an
⇒⧸(2k)⧸(2k)+23
⇒18
Q8 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 8
Write each expression such that there is no variable in the denominator .
(i) 3r−2s312r−3s7
Sol :
⇒(3r-2s3)(r3s-7)/12
Using property: am×an=a(m+n)
⇒312(r−2+3s3−7)
⇒14(r1s−4)
⇒r1s−44
(ii) 12x3y−2z46x7y−5z−3
Sol :
⇒126(x3y−2z4)(x−7y5z3)
Using property: am×an=a(m+n)
⇒2(x3-7×y-2+5×z4+3)
⇒2(x-4×y3×z7)
⇒2x-4y3z7
(iii) (5m)0n−24mn−3
Sol :
⇒504(m0×n−2)(m−1n3)
Using property: a0=1
⇒14(1×n−2)(m−1n3)
Using property: am×an=a(m+n)
⇒14(n−2+3)(m−1)
⇒nm−14
(iv) 14a−4k70a3k8
Sol :
⇒1470(a−4k)(a−3k−8)
Using property: a0=1
⇒14 (a-4k)(a-3k-8)
⇒14 (a-4-3k1-8)
⇒14 (a-7k-7)
Q9 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 9
Simplify:
(i) 2−n.82n−1.162n43n
Sol :
⇒2−n×23(2n−1)×24(2n)22(3n)
⇒2−n×26n−3×28n26n
⇒2−n×⧸(26n)×2−3×28n⧸(26n)
⇒2-n× 2-3× 28n
Using property : am×an=a(m+n)
⇒2-n-3+8n
⇒2+7n-3
(ii) 2n+4−2.2n2.2n+3+2−3
Sol :
Using property : am×an=a(m+n)
⇒2n+4−2n+12n+4+2−3
⇒⧸(2n+4)⧸(2n+4)−2n+12n+4+2−3
⇒1−2n+12(n+1)+(3)+2−3
⇒1−⧸(2n+1)⧸(2(n+1))×2(3)+2−3
⇒1−123+123
⇒1
(iii) (0.6)0−(0.1)−1(323)−1.(32)3+(−13)−1
Sol :
Using property: a0=1
⇒1−(110)−1(323)−1×(32)3+(−13)−1
Using property: a−1=1a
⇒1−101(233)×(3323)−(31)
⇒1−10(⧸(8)⧸(3))×(⧸(27)⧸(8))−3
⇒−99−3=−96=−32
Q10 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 10
If 9n×35×(27)33×(81)4=27 , then find the value of n
Sol :
⇒9n×35×(27)33×(81)4
⇒32n×35×3931×316
Using property: am×an=a(m+n)
⇒32n+5+9316+1
⇒32n+14317
⇒32n×314×3-17
⇒32n×314-17
⇒32n×3-3
⇒32n33=27 according to question
⇒32n=33×33
⇒32n=36
⇒32n=32×3
⇒n=3
Q11 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 11
Show that :
(i) (xaxb)(a+b)×(xbxc)(b+c)×(xcxa)(c+a)=1
Sol :
Using property : aman=a(m−n)
⇒{x(a-b)}(a+b) × {x(b-c)}(b+c) × {x(c-a)}(c+a)
Using property : (am)n=am×n
⇒x(a-b)×(a+b) × x(b-c)×(b+c) × x(c-a)×(c+a)
Using property : (a+b)(a-b)=a2-b2
⇒xa2−b2×xb2−c2×xc2−a2
Using property : am×an=a(m+n)
⇒xa2−b2+b2−c2+c2−a2
⇒x0
Using property : a0=1
⇒1
Hence proved
(ii) xa+b×xb+c×xc+a(xa×xb×xc)2=1
Sol :
Using property : am×an=a(m+n)
⇒xa+b+b+c+c+a(xa+b+c)2
⇒x2a+2b+2c(xa+b+c)2
⇒⧸(x2a+2b+2c)⧸(x2a+2b+2c)
⇒1
Hence proved
Q12 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 12
Find x such that
(i) (74)−3×(74)−5=(74)3x−2
Sol :
Using property : am×an=a(m+n)
⇒(74)−3−5=(74)3x−2
⇒(74)−8=(74)3x−2
On comparing both sides
⇒-8=3x-2
Transposing -2
⇒-8+2=3x
Transposing 3
⇒x=−63=−2
(ii) (1258)×(1258)x=(52)18
Sol :
Using property : am×an=a(m+n)
⇒(1258)x+1=(52)18
⇒(5323)x+1=(52)18
⇒(52)3(x+1)=(52)18
Comparing both sides
⇒3(x+1)=18
⇒3x+3=18
⇒3x=18-3
⇒x=153=5
Q13 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper
Question 13
Find the value of x such that
(i) 32x−1=127x−3
Sol :
⇒32x−1=133(x−3)
⇒32x-1=3-3x+9
On comparing both sides
⇒2x-1=-3x+9
⇒3x+2x=9+1
⇒5x=10
⇒x=105=2
(ii) (pq)3x+2=(qp)2−x
Sol :
⇒(pq)3x+2=(pq)−2+x
On comparing both sides
⇒3x+2=-2+x
⇒3x-x=-2-2
⇒2x=-4
⇒x=−42=−2
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